As usual, we split the operator \(I_\beta \)  in its local part and its global part
\(I_{\beta }f(x)=I_{\beta ,L}f(x)+I_{\beta ,G}f(x),\) 
where
\(I_{\beta ,L}f(x)=I_{\beta }(f\chi _{B_{h}(\cdot )} )(x)\) 
is the local part,
\(I_{\beta ,G}f(x)=I_{\beta }(f\chi _{B^{c}_{h}(\cdot )} )(x)\) 
is the global part, and for \(x\in R^{d}\)  by taking \(m(x)=1\wedge \frac{1}{|x|}\) ,
\(B_{h}(x):=\lbrace y\in R^{d}:|x-y|<dm(x)\rbrace \)  is an \(hiperbolic\)  \(ball\)  (\(admissible\)  \(ball\) ).
Let us take
\(\omega (s)=\displaystyle \frac{e^{-\frac{|y-e^{-s}x|^{2}}{1-e^{-2s}}}}{(1-e^{-2s})^{\frac{d}{2}}}\) ,
then, from (REF )
\(N_{\beta /2}(x,y)&=&\frac{1}{\pi ^{\frac{d}{2}}\Gamma (\frac{\beta }{2})}\int _{0}^{+\infty }t^{\frac{\beta }{2}-1}\left(\omega (t)-\omega (+\infty )\right)dt\\&=&\frac{1}{\pi ^{\frac{d}{2}}\Gamma (\frac{\beta }{2})}\int _{0}^{+\infty }t^{\frac{\beta }{2}-1}\left(-\int _{t}^{+\infty }\frac{\partial \omega (s)}{\partial s}ds\right)dt.\) 
Thus, using Hardy's inequality, see [1]}
\(\left|N_{\beta /2}(x,y)\right|\le \frac{1}{\pi ^{\frac{d}{2}}\Gamma (\frac{\beta }{2})}\int _{0}^{+\infty }t^{\frac{\beta }{2}-1}\int _{t}^{+\infty }\left|\frac{\partial \omega (s)}{\partial s}\right|ds\;dt \le \frac{1}{\pi ^{\frac{d}{2}}\Gamma (\frac{\beta }{2})}\frac{2}{\beta }\int _{0}^{+\infty }s^{\frac{\beta }{2}}\left|\frac{\partial \omega (s)}{\partial s}\right|ds.\) 
Now,
\(\frac{\partial \omega (s)}{\partial s}&=&\frac{(1-e^{-2s})^{\frac{d}{2}}e^{-\frac{|y-e^{-s}x|^{2}}{1-e^{-2s}}}}{(1-e^{-2s})^{d}}\left(\frac{-2(1-e^{-2s})(y-e^{-s}x)\cdot (e^{-s}x)+|y-e^{-s}x|^{2}e^{-2s}}{(1-e^{-2s})^{2}}\right)\\&&\hspace{128.0374pt}-\frac{1}{(1-e^{-2s})^{d}}\left(\frac{d}{2}e^{\frac{|y-e^{-s}x|^{2}}{1-e^{-2s}}}(1-e^{-2s})^{\frac{d}{2}-1}2e^{-2s}\right)\\&=&\omega (s)\left(\frac{-2(1-e^{-2s})(y-e^{-s}x)\cdot (e^{-s}x)+|y-e^{-s}x|^{2}e^{-2s}}{(1-e^{-2s})^{2}}-\frac{de^{-2s}}{(1-e^{-2s})}\right).\) 
Then, taking \(u=1-e^{-2s},\; du=2e^{-2s}ds,\)  i.e., \(e^{-s} = \sqrt{1-u},\)  we have
\(\left|N_{\beta /2}(x,y)\right|&\le &\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-\sqrt{1-u}x|^{2}}{u}}}{u^{\frac{d}{2}}}\\&&\hspace{5.69046pt}\times \left(\frac{2u|y-\sqrt{1-u}x|\sqrt{1-u}|x|+|y-\sqrt{1-u}x|^{2}(1-u)}{u^{2}}+\frac{d(1-u)}{u}\right)\frac{du}{2(1-u)}\\&=&\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-\sqrt{1-u}x|^{2}}{u}}}{u^{\frac{d}{2}}}\left(\frac{|y-\sqrt{1-u}x||x|}{u\sqrt{1-u}}+\frac{|y-\sqrt{1-u}x|^{2}}{2u^{2}}+\frac{d}{2u}\right)du\\&=& \int _{0}^{1}\frac{\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}}{u^{\frac{d}{2}+1}}e^{-\frac{|y-\sqrt{1-u}x|^{2}}{u}}\frac{|y-\sqrt{1-u}x||x|}{\sqrt{1-u}}du \\&& \hspace{28.45274pt}+\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-\sqrt{1-u}x|^{2}}{u}}}{u^{\frac{d}{2}+1}}\frac{|y-\sqrt{1-u}x|^{2}}{2u}du\\&& \hspace{42.67912pt}+\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-\sqrt{1-u}x|^{2}}{u}}}{u^{\frac{d}{2}}}\frac{d}{2u} du\\&&=I+II+III,\) 
where
\(I = \int _{0}^{1}\frac{\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}}{u^{\frac{d}{2}+1}}e^{-\frac{|y-\sqrt{1-u}x|^{2}}{u}}\frac{|y-\sqrt{1-u}x||x|}{\sqrt{1-u}}du,\) 
\(II= \int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-\sqrt{1-u}x|^{2}}{u}}}{u^{\frac{d}{2}+1}}\frac{|y-\sqrt{1-u}x|^{2}}{2u}du,\) 
and
\(III = \int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-\sqrt{1-u}x|^{2}}{u}}}{u^{\frac{d}{2}}}\frac{d}{2u} du.\) 

Let us study the local part first. We need to bound each of the terms \(I\) , \(II\)  and \(III\)  in this part.
For \(I\) , since we are in the local part
\(|x-y| \le \frac{d}{|x|}\) , then we have \(|x-y||x| \le d \)  therefore,
\(\nonumber |y-\sqrt{1-u}\,x|^{2} & \ge & (|y-x|-|x|(1-\sqrt{1-u}))^{2} \\& \ge & |y-x|^{2}-2|x||y-x| \frac{u}{1+\sqrt{1-u}} \ge |y-x|^{2}-2 d \; u.\) 
On the other hand, it is well known that, there exist \(C>0\)  such that
for any \(x>0\) , \(\alpha \ge 0\)  and \(c>0\) ,
\(x^\alpha e^{-cx^2} \le C \hspace{14.22636pt}\cdot \) 
Thus, using (REF ) and (REF ) twice, with \(\alpha =1,\; c= \frac{1}{2}\)  and \(\alpha =d-1, c=\frac{1}{2}\) , we get
\(I&=&\int _{0}^{1}\frac{\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}}{u^{\frac{d}{2}}}e^{-\frac{|y-\sqrt{1-u}x|^{2}}{2u}}\left(\left|\frac{y-\sqrt{1-u}x}{\sqrt{u}}\right|e^{-\frac{|y-\sqrt{1-u}x|^{2}}{2u}}\right)\frac{|x|}{\sqrt{u}\sqrt{1-u}}du\\&\le &C|x|\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-x|^{2}}{2u}}}{u^{\frac{d}{2}}}\frac{du}{\sqrt{u}\sqrt{1-u}}\) 
\(&=&C|x|\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-x|^{2}}{2u}}}{u^{\frac{d-1}{2}}}\frac{du}{u\sqrt{1-u}}\\&\le &\frac{C|x|}{|x-y|^{d-1}}\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{du}{u\sqrt{1-u}}.\\\) 
By Lemma REF  we get
\(&&\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{du}{u\sqrt{1-u}}\\&& \hspace{28.45274pt} = \int _{0}^{1/2}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{du}{u\sqrt{1-u}}\; +\;\int _{1/2}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{du}{u\sqrt{1-u}}\\&& \hspace{28.45274pt} \le C\int _{0}^{1/2}\frac{\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}}{u}du\; +\;C\int _{1/2}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{du}{\sqrt{1-u}}<\infty ,\) 
since \(\displaystyle \frac{1}{u}\)  in bounded on \([1/2,1]\)  and
\(\displaystyle \frac{1}{\sqrt{1-u}}\)  is bounded on \([0,1/2]\) . Thus, we have
\(I\le \frac{C|x|}{|x-y|^{d-1}}.\) 
For \(II\) , we use again (REF ) and (REF ) with \(\alpha =2\)  and \(c=\frac{1}{2}\)  we have
\(II &=&\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-\sqrt{1-u}x|^{2}}{u}}}{u^{\frac{d}{2}+1}}\frac{|y-\sqrt{1-u}x|^{2}}{2u}du\\&\le &C\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-\sqrt{1-u}x|^{2}}{2u}}}{u^{\frac{d}{2}+1}}du\\&\le & C\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-x|^{2}}{2u}}}{u^{\frac{d}{2}+1}}du\\&=&C \int _{0}^{1/2}\frac{\left(-log(\sqrt{1-u})\right)}{u}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}-1}\frac{e^{-\frac{|y-x|^{2}}{2u}}}{u^{\frac{d}{2}}}du\;+\;\int _{1/2}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-x|^{2}}{2u}}}{u^{\frac{d}{2}+1}}du.\\\) 
Since \(\displaystyle \lim _{u\rightarrow 0}\frac{\left(-log(\sqrt{1-u})\right)}{u}=\frac{1}{2}\) , this function is bounded on \([0,1/2]\)  and
\(\displaystyle \frac{1}{u}\)  is bounded on \([1/2,1]\) , thus we get
\(II&\le &C\int _{0}^{1/2}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}-1}\frac{e^{-\frac{|y-x|^{2}}{2u}}}{u^{\frac{d}{2}}}du\;+\;C\int _{1/2}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-x|^{2}}{2u}}}{u^{\frac{d}{2}}}du\\&\le &C\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}-1}\frac{e^{-\frac{|y-x|^{2}}{2u}}}{u^{\frac{d}{2}}}du\;+\;C\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-x|^{2}}{2u}}}{u^{\frac{d}{2}}}du\\&=&C\mathcal {K}_{2}(x-y)\;+\;CG_{2}(x-y),\) 
where
\( \mathcal {K}_2(x) := \int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}-1}\frac{e^{-\frac{|x|^{2}}{2u}}}{u^{\frac{d}{2}}}du,\) 
and
\(G_{2}(x):=\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|x|^{2}}{2u}}}{u^{\frac{d-1}{2}}}\frac{du}{\sqrt{u}}.\) 
Again by (REF )
\(G_{2}(x-y)&=&\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-x|^{2}}{2u}}}{u^{\frac{d-1}{2}}}\frac{du}{\sqrt{u}}\\&\le & \frac{C}{|x-y|^{d-1}}\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{du}{\sqrt{u}}.\) 
Now, by Hölder's inequality with \(p=\frac{3}{2}\)  y \(q=3\)  we have
\(\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{du}{\sqrt{u}}\le \left(\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{3\beta }{2}}du\right)^{1/3}\left(\int _{0}^{1}\frac{du}{u^{3/4}}\right)^{2/3}.\) 
Thus, by Lemma (REF ) we obtain
\(\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{3\beta }{2}}du=\int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{(\frac{3\beta }{2}+1)-1}du\;<\;\infty \) 
and trivially
\(\displaystyle {\int _{0}^{1}\frac{du}{u^{3/4}}}< \infty .\)
Finally for \(III\) , by analogous arguments as in \(II\) , we get
\(III = \int _{0}^{1}\left(-log(\sqrt{1-u})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{|y-\sqrt{1-u}x|^{2}}{u}}}{u^{\frac{d}{2}}}\frac{d}{2u} du\le C \mathcal {K}_{2}(x-y)\;+\;CG_{2}(x-y).\) 
Therefore,
\(\left|N_{\beta /2}(x,y)\right|&\le & I+II+III\\&\le & \frac{C|x|}{|x-y|^{d-1}}\,+\,C\mathcal {K}_{2}(x-y)\,+\,\frac{C}{|x-y|^{d-1}}\\&\le & C\frac{|x|+1}{|x-y|^{d-1}}\,+\,C\mathcal {K}_{2}(x-y)= C\mathcal {K}_{3}(x,y)\,+\,C\mathcal {K}_{2}(x-y),\) 
where
\( \mathcal {K}_{3}(x,y) := \frac{|x|+1}{|x-y|^{d-1}}.\) 
Thus, using the above estimates, we conclude that the local part
\(I_{\beta ,L}\)  can be bounded as
\(|I_{\beta ,L} f(x)| &=& |I_{\beta } (f\chi _{B_h(\cdot )})(x)| = \Big | \int _{B_h(x)} N_{\beta /2} (x,y) f(y) \;dy \Big |\\&\lesssim & \int _{B_h(x)} {\mathcal {K}}_3(x,y) |f(y)| \;dy + \int _{B_h(x)} {\mathcal {K}}_2(x-y) | f(y) |\;dy \\&=& IV + V.\\\) 
Now, to bound IV and V, we need to take a countable family of admissible balls \( \mathcal {F}\)  that satisfies the condition of Lemma 4.3 of [2]}. In particular, \(\mathcal {F}\)  verifies

For each \(B \in \mathcal {F}\)  let \(\tilde{B}=2 B,\)  then, the family of those balls \(\tilde{\mathcal {F}}= \lbrace B(0,1),\lbrace \tilde{B}\rbrace _{B \in \mathcal {F}}\rbrace \)  is a covering of \(\mathbb {R}^{d}\) ;

\(\mathcal {F}\)  has a bounded overlaps property;

Every ball \(B \in \mathcal {F}\)  is contained in an admissible ball, and therefore for
any pair \(x, y \in B, e^{-|x|^{2}} \sim e^{-|y|^{2}}\)  with constants independent of \(B\)

There exists a uniform positive constant \(C_{d}\)  such that, if \(x \in B \in \mathcal {F}\)  then \(B_h(x) \subset C_{d} B:=\hat{B} .\)  Moreover, the collection \(\hat{\mathcal {F}}=\lbrace \hat{B}\rbrace _{B \in \mathcal {F}}\)  also satisfies the properties ii) and iii).

Given \(B \in \mathcal {F},\)  if \(x \in B\)  then \(B_h(x) \subset \hat{B}\) , we get,
\(IV &=& (1+|x|) \sum _{k=0}^\infty \int _{2^{-(k+1)}C_d m(x) < |x-y| < 2^{-k}C_d m(x)} \frac{|f(y)| \chi _ {\hat{B}}}{|x-y|^{d-1}} dy\\&\le & C_d 2^d \mathcal {M}(f\chi _{\hat{B}})(x) (1+|x|) m(x)\sum _{k=0}^\infty 2^{-(k+1)} \le C\mathcal {M}(f\chi _{\hat{B}})(x)\chi _{B}(x),\) 
where \(\mathcal {M}(g)\)  is the classical Hardy-Littlewood maximal
function of the function \(g.\)
On the other hand, let us consider the function
\(\varphi (y) =\frac{1}{\pi ^{d/2}} e^{-\frac{1}{2}|y|^2},\) 
then \(\displaystyle \int _{\mathbb {R}^d} \varphi (y) dy =1\) . It is well known that
\(\varphi \)  is a non-increasing radial function, and given \(t>0,\,\) 
we rescale this function as \(\varphi _{\sqrt{t}} (y) = t^{-d/2}\varphi (y/\sqrt{t}),\)  and, since \(0\le \varphi \in L^1(\mathbb {R}^d),\)  \(\left\lbrace \varphi _{\sqrt{t}}\right\rbrace _{t>0}\;\)  is the
classical (Gauss-Weiertrass) approximation of the identity in \(\mathbb {R}^d.\,\)  Then,
since
\(\displaystyle \int _{0}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}-1}dt <\infty ,\) 
we get
\(V &=& \int _{B_h(x)} \mathcal {K}_2(x-y) | f(y) |\;dy =\int _{B_h(x)} \Big (\int _0^1 \varphi _{\sqrt{t}}(x-y)\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}-1}dt\Big ) |f(y)| dy\\&\le & \int _{B_h(x)} \Big (\sup _{t>0} \varphi _{\sqrt{t}}(x-y)\Big )\Big (\int _0^1\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}-1}dt\Big ) |f(y)| dy\\&\le & C \int _{B_h(x)} \Big (\sup _{t>0} \varphi _{\sqrt{t}}(x-y)\Big ) |f(y)| dy.\) 
Again, using the family \(\mathcal {F}\) , if \(x \in B\)  then \(B_h(x)\subset \hat{B}\) . By a similar argument as before and as in as result in Stein's book [1]},
\(V &=& \int _{B_h(x)} \mathcal {K}_2(x-y) | f(y) |\;dy \le C\int _{\mathbb {R}^d} \left(\sup _{t>0} \varphi _{\sqrt{t}}(x-y) \right)|f(y)| \chi _{\hat{B}}(y) dy\\&\lesssim & \sum _{B \in \mathcal {F}} \left| \left( \sup _{t>0}\varphi _{\sqrt{t}} * |f \chi _{\hat{B}}|\right)(x) \right| \chi _{B}(x)\le \sum _{B \in \mathcal {F}} \mathcal {M}(f \chi _{\hat{B}})(x)\chi _{B}(x),\) 
which yields,
\(|I_{\beta ,L}f(x)|\le \displaystyle \sum _{B \in \mathcal {F}} \mathcal {M}(f \chi _{\hat{B}})(x)\chi _{B}(x).\)
Then, for \(f \in L^{p(\cdot )}\left(\mathbb {R}^{d}, \gamma _d\right)\)  we will use the characterization of the norm by duality, we get
\(\left\Vert I_{\beta ,L}(f)\right\Vert _{p(\cdot ), \gamma _{d}} \le 2 \sup _{\Vert g\Vert _{p^{\prime }(\cdot ), \gamma _{d}} \le 1} \int _{\mathbb {R}^{d}}\left|I_{\beta ,L}f(x)\right||g(x)| \gamma _d(dx).\) 
Using the estimates above, we get
\(\int _{\mathbb {R}^{d}}\left|I_{\beta ,L}f(x)\right||g(x)| \gamma _d(dx) &\lesssim &\sum _{B \in \mathcal {F}} \int _{B} \mathcal {M}\left(f \chi _{\hat{B}(\cdot )}\right)(x)|g(x)| e^{-|x|^{2}} d x \\&& \approx \sum _{B \in \mathcal {F}} e^{-\left|c_{B}\right|^{2}} \int _{B} \mathcal {M}\left(f \chi _{\hat{B}(\cdot )}\right)(x)|g(x)| d x,\) 
where \(c_{B}\)  is the center of \(B\)  and \(\hat{B}\)  and we have used property iii) above, i.e. that over each ball of the family \(\mathcal {F},\)  the values of \(\gamma _d\)  are all equivalent.
Applying Hölder's inequality for \(p(\cdot )\)  and \(p^{\prime }(\cdot )\)  with respect of the Lebesgue measure and the boundedness of \(\mathcal {M}\)  on \(L^{p(\cdot )}\left(\mathbb {R}^{d}\right),\)  we get
\(\nonumber \int _{\mathbb {R}^{d}}\left|I_{\beta ,L}f(x)\right||g(x)| \gamma _{d}(d x) &\lesssim & \sum _{B \in \mathcal {F}} e^{-\left|c_{B}\right|^{2}}\left\Vert \mathcal {M}\left(f \chi _{\hat{B}(\cdot )}\right) \chi _{B}\right\Vert _{p(\cdot )}\left\Vert g \chi _{B}\right\Vert _{p^{\prime }(\cdot )} \\\nonumber &\lesssim & \sum _{B \in \mathcal {F}} \mathrm {e}^{-\left|c_{B}\right|^{2}}\left\Vert f \chi _{\hat{B}}\right\Vert _{p(\cdot )}\left\Vert g \chi _{B}\right\Vert _{p^{\prime }(\cdot )} \\&=& \sum _{B \in \mathcal {F}} \mathrm {e}^{-\left|c_{B}\right|^{2} / p_{\infty }}\left\Vert f \chi _{\hat{B}}\right\Vert _{p(\cdot )} e^{-\left|c_{B}\right|^{2} / p_{\infty }^{\prime }}\left\Vert g \chi _{\hat{B}}\right\Vert _{p^{\prime }(\cdot )}.\) 
Since \(p \in P_{\gamma _{d}}^{\infty }\left(\mathbb {R}^{d}\right)\)  and \(p_{-}>1\)  then \(p^{\prime }\in P_{\gamma _{d}}^{\infty }\left(\mathbb {R}^{d}\right) .\)  Thus, from Lemma \(1.4,\)  for
every \(x \in \mathbb {R}^{d}\)
\(e^{-|x|^{2}\left(p(x) / p_{\infty }-1\right)} \le C_{1} \text{ and } e^{-|x|^{2}\left(p^{\prime }(x) / p_{\infty }^{\prime }-1\right)} \le C_{2}.\) 
Moreover, since the values of the Gaussian measure \(\gamma _{d}\)  are all equivalent on any ball \(\hat{B} \in \tilde{\mathcal {F}}\) , we have
\(\int _{\hat{B}}\left(\frac{|f(y)|}{e^{\left|c_{B}\right|^{2} / p_{\infty }}\left\Vert f \chi _{\hat{B}}\right\Vert _{p(\cdot ), \gamma _{d}}}\right)^{p(y)} d y &\lesssim & \int _{\hat{B}}\left(\frac{|f(y)|}{\left\Vert f \chi _{\hat{B}}\right\Vert _{p(\cdot ), \gamma _{d}}}\right)^{p(y)} e^{-|y|^{2}\left(p(y) / p_{\infty }-1\right)} \gamma _{d}(dy) \\&\lesssim &\int _{\hat{B}}\left(\frac{|f(y)|}{\left\Vert f \chi _{\hat{B}}\right\Vert _{p(\cdot ), \gamma _{d}}}\right)^{p(y)} \gamma _{d}(dy) \lesssim 1,\) 
which yields
\(e^{-\left|c_{B}\right|^{2} / p_{\infty }}\left\Vert f \chi _{\hat{B}}\right\Vert _{p(\cdot )} \lesssim \left\Vert f \chi _{\hat{B}}\right\Vert _{p(\cdot ), \gamma _{d}}.\) 
Similarly, by the second inequality of (REF ) we also get
\(e^{-\left|c_{B}\right|^{2} / p_{\infty }^{\prime }}\left\Vert g \chi _{\hat{B}}\right\Vert _{p^{\prime }(\cdot )} \lesssim \left\Vert g \chi _{\hat{B}}\right\Vert _{p^{\prime }(\cdot ), \gamma _{d}}.\) 
Replacing both estimates in (REF ) we obtain
\(\int _{\mathbb {R}^{d}}\left|I_{\beta ,L}f(x)\right||g(x)| \gamma _{d}(d x) & \lesssim &\sum _{B \in \mathcal {F}}\left\Vert f \chi _{\hat{B}}\right\Vert _{p(\cdot ), \gamma _{d}}\left\Vert g \chi _{\hat{B}}\right\Vert _{p^{\prime }(\cdot ), \gamma _{d}} \\&=& \sum _{B \in \mathcal {F}}\left\Vert f \chi _{\hat{B}} e^{-|\cdot |^{2} / p(\cdot ) |}\right\Vert _{p(\cdot )}\left\Vert g \chi _{\hat{B}} e^{-|\cdot |^{2} / p^{\prime }(\cdot )}\right\Vert _{p^{\prime }(\cdot )}.\) 
Since the family of balls \(\hat{\mathcal {F}}\)  has bounded overlaps, from Corollary 1.1 applied to \(f e^{-|\cdot |^{2} / p(\cdot )} \in L^{p(\cdot )}(\mathbb {R}^{d})\)  and \(g e^{-|\cdot |^{2} / p^{\prime }(\cdot )} \in L^{p^{\prime }(\cdot )}(\mathbb {R}^{d}),\)  it follows that
\(\int _{\mathbb {R}^{d}}\left|I_{\beta ,L}f(x)\right||g(x)| \gamma _{d}(d x) \leqslant \Vert f\Vert _{p(\cdot ), \gamma _{d}}\Vert g\Vert _{p^{\prime }(\cdot ), \gamma _{d}}.\) 
Taking supremum over all functions \(g\)  with \(\Vert g\Vert _{p^{\prime }(\cdot ), \gamma _{d}} \le 1,\)  from (REF ) we get finally
\(\left\Vert I_{\beta ,L}(f)\right\Vert _{p(\cdot ), \gamma _{d}}\le C\Vert f\Vert _{p(\cdot ), \gamma _{d}}.\\\) 
Therefore, the local part \(I_{\beta ,L}\)  is bounded in \(L^{p(\cdot )}(\gamma _d)\) .

Now, let us study the global part. Again, since
\(\left|N_{\beta /2}(x,y)\right|\le I+II+III,\) 
we need to estimate each term in this part. As usual for the global part, the arguments are completely different but are based on the following technical result, obtained S. Pérez [4]}, see also [2]}.To simplify the notation, in what follows we denote
\(a=a(x, y):=|x|^{2}+|y|^{2}, b=b(x, y):=2\langle x, y\rangle \text{, }\) 
\(u(t)=u(t ; x, y):=\frac{|y-\sqrt{1-t x}|^{2}}{t}=\frac{a}{t}-\frac{\sqrt{1-t}}{t} b-|x|^{2},\) 
\(t_{0}=2 \frac{\sqrt{a^{2}-b^{2}}}{a+\sqrt{a^{2}-b^{2}}},\) 
then
\(u\left(t_{0}\right)=\frac{\sqrt{a^{2}-b^{2}}}{2}+\frac{a}{2}-|x|^{2}=\frac{|y|^{2}-|x|^{2}}{2}+\frac{\sqrt{a^{2}-b^{2}}}{2}\) 
and
\(t_{0} \sim \frac{\sqrt{a^{2}-b^{2}}}{a} \sim \frac{\sqrt{a-b}}{\sqrt{a+b}}=\frac{|x-y|}{|x+y|}.\) 
It is well known that \(t_{0}<1\) , the minimum of \(u(t)\)  is attained at
\(t_{0}\)  and
\(\frac{1}{t_{0}^{d/2}}\lesssim |x+y|^{d}.\) 
For details and other properties of these terms, see [4]}, [7]} or [2]}.
Let us fix \(x\in R^{d}\)  and consider \(E_{x}=\lbrace y\in R^{d}:b>0\rbrace \) .

Case \(b\le 0\):
First, for \(0< \epsilon <1\) , using inequality (REF ) we have
\(I&=&\int _{0}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-u(t)}}{t^{\frac{d}{2}}}\frac{\left|y-\sqrt{1-t}x\right||x|}{t\sqrt{1-t}}dt\\&=&|x|\int _{0}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-\epsilon u(t)-(1-\epsilon )u(t)}}{t^{\frac{d+1}{2}}}\frac{\left|y-\sqrt{1-t}x\right|}{\sqrt{t}\sqrt{1-t}}dt\\&\le &C_{\epsilon }|x|\int _{0}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )u(t)}}{t^{\frac{d+1}{2}}}\frac{dt}{\sqrt{1-t}}.\) 
Since
\(\displaystyle \frac{\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}}{\sqrt{1-t}}\) 
is continuous on \([0,1/2]\)  and therefore bounded, we get
\(\int _{0}^{1/2}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )u(t)}}{t^{\frac{d+1}{2}}}\frac{dt}{\sqrt{1-t}}\le C_{\beta }\int _{0}^{1/2}\frac{e^{-(1-\epsilon )u(t)}}{t^{\frac{d+1}{2}}}dt,\) 
and since \(0<t<1/2\) , \(\frac{1}{\sqrt{t}}<\frac{1}{t}\)  and then, by (REF ), we get
\(\int _{0}^{1/2}\frac{e^{-(1-\epsilon )u(t)}}{t^{\frac{d+1}{2}}}dt&\le &e^{(1-\epsilon )|x|^{2}}\int _{0}^{1/2}\frac{e^{-(1-\epsilon )\frac{a}{t}}}{t^{\frac{d+1}{2}}}dt\\&\le &e^{(1-\epsilon )|x|^{2}}\int _{0}^{1/2}\frac{e^{-(1-\epsilon )\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt\\&\le &e^{(1-\epsilon )|x|^{2}}C_{\epsilon }e^{-(1-\epsilon )a}= C_{\epsilon }e^{-(1-\epsilon )|y|^{2}}.\) 
Analogously,
\(\int _{1/2}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )u(t)}}{t^{\frac{d+1}{2}}}\frac{dt}{\sqrt{1-t}}&\le &e^{(1-\epsilon )|x|^{2}}\int _{1/2}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )\frac{a}{t}}}{t^{\frac{d+1}{2}}}\frac{dt}{\sqrt{1-t}}\\&\le &C_{d}e^{(1-\epsilon )|x|^{2}}\int _{1/2}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )\frac{a}{t}}}{\sqrt{1-t}}dt,\) 
since \(\displaystyle \frac{1}{t^{\frac{d+1}{2}}}\)  is bounded on
\([1/2,1]\) , also as \(1/2<t<1\)  we have \(-\frac{a}{t}<-a\)  and then by Lemma  REF
\(\int _{1/2}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )\frac{a}{t}}}{\sqrt{1-t}}dt&\le &e^{-(1-\epsilon )a}\int _{1/2}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{dt}{\sqrt{1-t}}\\&=&C_{\beta }e^{-(1-\epsilon )a}.\) 
Thus,
\(\int _{1/2}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )u(t)}}{t^{\frac{d+1}{2}}}\frac{dt}{\sqrt{1-t}}\le C_{d,\beta }e^{(1-\epsilon )|x|^{2}}e^{-(1-\epsilon )a}=Ce^{-(1-\epsilon )|y|^{2}}.\) 
Therefore
\(I \le C_\epsilon |x| e^{-(1-\epsilon )|y|^{2}}.\) 
Now, using again inequality (REF ), we get
\(II&=&\int _{0}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-u(t)}}{t^{\frac{d}{2}}}\frac{\left|y-\sqrt{1-t}x\right|^{2}}{2t^{2}}dt\\&=&\frac{1}{2}\int _{0}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-\epsilon u(t)-(1-\epsilon )u(t)}}{t^{\frac{d}{2}+1}}\frac{\left|y-\sqrt{1-t}x\right|^{2}}{t}dt\\&\le &C_{\epsilon }\int _{0}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )u(t)}}{t^{\frac{d}{2}+1}}dt\\&\le &C_{\epsilon }e^{(1-\epsilon )|x|^{2}}\int _{0}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt.\\\) 
Now,
\(\int _{0}^{1/2}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt\le C\int _{0}^{1/2}\frac{e^{-(1-\epsilon )\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt\le C\int _{0}^{1}\frac{e^{-(1-\epsilon )\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt.\) 
Therefore, by taking the change of variables \(s=a(\frac{1}{t}-1)\) , we get
\(\nonumber \int _{0}^{1}\frac{e^{-(1-\epsilon )\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt&=&\int _{0}^{+\infty }e^{-(1-\epsilon )(s+a)}\left(\frac{s+a}{a}\right)^{\frac{d}{2}+1}\frac{a}{\left(s+a\right)^{2}}ds\\\nonumber &=&\frac{e^{-(1-\epsilon )a}}{a^{\frac{d}{2}}}\int _{0}^{+\infty }e^{-(1-\epsilon )s}\left(s+a\right)^{\frac{d}{2}-1}ds\\\nonumber &\le &\frac{Ce^{-(1-\epsilon )a}}{a^{\frac{d}{2}}}\int _{0}^{+\infty }e^{-(1-\epsilon )s}\left(s^{\frac{d}{2}-1}+a^{\frac{d}{2}-1}\right)ds\\\nonumber &\le &\frac{C_{\epsilon }e^{-(1-\epsilon )a}}{a^{\frac{d}{2}}}\left(\Gamma \left(\frac{d}{2}\right)\;+\;a^{\frac{d}{2}-1}\right)\\\nonumber &=&C_{\epsilon }e^{-(1-\epsilon )a}\left(\frac{\Gamma \left(\frac{d}{2}\right)}{a^{\frac{d}{2}}}\;+\;\frac{1}{a}\right)\\&\le &C_{\epsilon }e^{-(1-\epsilon )a},\) 
since \(a\ge \frac{d}{2}\) . Analogously, by Lemma  REF
\(\int _{1/2}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )}}{t^{\frac{d}{2}+1}}dt&\le &C\int _{ 1/2}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}e^{-(1-\epsilon )\frac{a}{t}}dt\\&\le &C\int _{1/2}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}e^{-(1-\epsilon )a}dt\\&\le &Ce^{-(1-\epsilon )a}\int _{0}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}dt \\&=& Ce^{-(1-\epsilon )a}.\) 
Then,
\(II\le C_{\epsilon }e^{(1-\epsilon )|x|^{2}}e^{-(1-\epsilon )a}=C_{\epsilon }e^{(1-\epsilon )|x|^{2}}e^{-(1-\epsilon )\left(|y|^{2}+|x|^{2}\right)}=C_{\epsilon }e^{-(1-\epsilon )|y|^{2}}.\) 
Finally,
\(III&=&\int _{0}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-u(t)}}{t^{\frac{d}{2}}}\frac{d}{2}\frac{dt}{t}\le \frac{d}{2}\int _{0}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}e^{-\frac{a}{t}+|x|^{2}}\frac{dt}{t^{\frac{d}{2}+1}}\\&= &Ce^{|x|^{2}}\int _{0}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt.\\\) 
Now,
\(\int _{0}^{1/2}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt\le C\int _{0}^{1/2}\frac{e^{-\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt\le C\int _{0}^{1}\frac{e^{-\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt\) 
since \(\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\)  is
bounded in \([0,1/2],\)  by taking
\(s=a(\frac{1}{t}-1), \;ds=-\frac{a}{t^{2}}dt,\) 
we get
\(\int _{0}^{1}\frac{e^{-\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt&=&\int _{0}^{+\infty }e^{-(s+a)}\left(\frac{s+a}{a}\right)^{\frac{d}{2}+1}\left(\frac{a}{s+a}\right)^{2}\frac{ds}{a}\\&=&\frac{e^{-|y|^{2}-|x|^{2}}}{a^{\frac{d}{2}}}\int _{0}^{+\infty }e^{-s}\left(s+a\right)^{\frac{d}{2}-1}ds.\\\) 
On the other hand,
\(\left(s+a\right)^{\frac{d}{2}-1}=\frac{\left(s+a\right)^{\frac{d}{2}}}{s+a}\le C\frac{\left(s^{\frac{d}{2}}+a^{\frac{d}{2}}\right)}{s+a}\le C\left(\frac{s^{\frac{d}{2}}}{s}+\frac{a^{\frac{d}{2}}}{a}\right)=C\left(s^{\frac{d}{2}-1}+a^{\frac{d}{2}-1}\right).\) 
Therefore,
\(\int _{0}^{1}\frac{e^{-\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt&\le &C\frac{e^{-|y|^{2}-|x|^{2}}}{a^{\frac{d}{2}}}\left(\int _{0}^{+\infty }e^{-s}s^{\frac{d}{2}-1}ds\;+\;\int _{0}^{+\infty }e^{-s}a^{\frac{d}{2}-1}ds\right)\\&=&C\frac{e^{-|y|^{2}-|x|^{2}}}{a^{\frac{d}{2}}}\left(\Gamma \left(\frac{d}{2}\right)\;+\;a^{\frac{d}{2}-1}\right)=Ce^{-|y|^{2}-|x|^{2}}\left(\frac{\Gamma \left(\frac{d}{2}\right)}{a^{\frac{d}{2}}}\;+\;\frac{1}{a}\right).\) 
Thus,
\(\int _{0}^{1}\frac{e^{-\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt\le Ce^{-|y|^{2}-|x|^{2}},\hspace{19.91684pt}\mbox{as }a\ge \frac{d}{2}.\) 
For \(\frac{1}{2}<t<1,\hspace{8.5359pt}-a>-\frac{a}{t}>-2a\) . Hence, by Lemma  REF
\(\int _{1/2}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-\frac{a}{t}}}{t^{\frac{d}{2}+1}}dt&\le &C\int _{1/2}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}e^{-a}dt\\&\le &Ce^{-a}\int _{0}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}dt\\&=&Ce^{-|y|^{2}-|x|^{2}}.\) 
Then,
\(III\le Ce^{|x|^{2}}e^{-|y|^{2}-|x|^{2}}= Ce^{-|y|^{2}}\le Ce^{-(1-\epsilon )|y|^{2}}.\) 
In other words,
\(I\le C_{\epsilon }|x|e^{-(1-\epsilon )|y|^{2}}\) 
and
\(II, \; III\le Ce^{-|y|^{2}}\le Ce^{-(1-\epsilon )|y|^{2}}.\) 
Thus, for \(b\le 0\)
\(\left|N_{\beta /2}(x,y)\right|\le C_{\epsilon }(|x|+1)e^{-(1-\epsilon )|y|^{2}}.\) 
Next, we take \(0<\epsilon <1/p^{\prime }_{-}\)  and \(\tilde{\epsilon }=1/p^{\prime }_{-} - \epsilon = 1-\epsilon -1/p_{-}\) . Then, \(\tilde{\epsilon }>0\)  and \(1-\epsilon =\tilde{\epsilon }+1/p_{-}\)  .Therefore, for \(f\in L^{p(\cdot )}(\mathbb {R}^{d},\gamma _{d})\)  with \(\Vert f\Vert _{p,\gamma _{d}}=1\) , using Hölder's inequality
\(\int \limits _{\mathbb {R}^{d}}\left(\int \limits _{B_{h}^{c}(x)\cap E^{c}_{x}}\left|N_{\beta /2}(x,y)\right||f(y)|dy\right)^{p(x)}\gamma _{d}(dx) &\lesssim &\int \limits _{\mathbb {R}^{d}}\left(\int \limits _{\mathbb {R}^{d}}(|x|+1)e^{-(1-\epsilon )|y|^{2}}|f(y)|dy\right)^{p(x)}\gamma _{d}(dx)\\ &=&\int \limits _{\mathbb {R}^{d}}\left(\int \limits _{\mathbb {R}^{d}}e^{-(\tilde{\epsilon }+1/p_{-})|y|^{2}}|f(y)|dy\right)^{p(x)}(|x|+1)^{p(x)}\gamma _{d}(dx)\\ &\le &\int \limits _{\mathbb {R}^{d}}\left(\int \limits _{\mathbb {R}^{d}}|f(y)|^{p_{-}}e^{-|y|^{2}}dy\right)^{p(x)/p_{-}}\left(\int \limits _{\mathbb {R}^{d}}e^{-\tilde{\epsilon }|y|^{2}p^{\prime }_{-}}dy\right)^{p(x)/p^{\prime }_{-}}\\&&\hspace{156.49014pt} \times (|x|+1)^{p_{+}}\gamma _{d}(dx)\\ &\lesssim &\int \limits _{\mathbb {R}^{d}}\left(\int \limits _{\mathbb {R}^{d}}|f(y)|^{p_{-}}e^{-|y|^{2}}dy\right)^{\frac{p(x)}{p_{-}}}(|x|+1)^{p_{+}}\gamma _{d}(dx),\) 
and, since \(\rho _{p(\cdot ),\gamma _{d}}(f)\le 1\) ,
\(\int \limits _{\mathbb {R}^{d}}|f(y)|^{p_{-}}e^{-|y|^{2}}dy&\lesssim &\int \limits _{|f|\ge 1}|f(y)|^{p_{-}}\gamma _{d}(dy)+\int \limits _{|f|<1}|f(y)|^{p_{-}}\gamma _{d}(dy)\\&\le &\int \limits _{|f|\ge 1}|f(y)|^{p(y)}\gamma _{d}(dy)+1\le \rho _{p(\cdot ),\gamma _{d}}(f)+1\le 2.\) 
Thus,
\(\int \limits _{\mathbb {R}^{d}}\left(\int \limits _{B_{h}^{c}(x)\cap E^{c}_{x}}\left|N_{\beta /2}(x,y)\right||f(y)|dy\right)^{p(x)}\gamma _{d}(dx) \lesssim 2^{\frac{p_{+}}{p_{-}}}\int \limits _{\mathbb {R}^{d}}(|x|+1)^{p_{+}}\gamma _{d}(dx)=C_{d,p}.\) 
Hence,
\(\Vert I_{\beta }(f\chi _{B_{h}^{c}(\cdot )\cap E^{c}_{(\cdot )}})\Vert _{p(\cdot ),\gamma _{d}}\le C_{d,p}.\) 

Case \(b>0\):
In this case, \(I\)  is a very problematic term so we will discuss it at the end.
Again, by inequality (REF )
\(II&=&\frac{1}{2}\int ^{1}_{0}\left(-log(\sqrt{1-t})\right)^{\beta /2}\frac{e^{-u(t)}}{t^{d/2+1}}\frac{\left|y-\sqrt{1-t}x\right|^{2}}{t}\,dt\\&=&\frac{1}{2}\int ^{1}_{0}\left(-log(\sqrt{1-t})\right)^{\beta /2}\frac{e^{-(1-\epsilon )u(t)}}{t^{d/2+1}} e^{-\epsilon u(t)}\frac{\left|y-\sqrt{1-t}x\right|^{2}}{t}\,dt\\&\le &C_{\epsilon }\int ^{1}_{0}\left(-log(\sqrt{1-t})\right)^{\beta /2}\frac{e^{-(1-\epsilon )u(t)}}{t^{d/2+1}}\,dt.\) 
By using inequality (4.44) of [2]}, see also [4]}, we get
\(\frac{e^{-u(t)}}{t^{d/2}}\le 2^{d}\frac{e^{-u(t_{0})}}{t_{0}^{d/2}}.\) 
Thus,
\(\nonumber \frac{e^{-(1-\epsilon )u(t)}}{t^{d/2}}&=&\left(\frac{e^{-u(t)}}{t^{d/2}}\right)^{1-\epsilon } \frac{1}{t^{\epsilon d/2}}\\&\lesssim &\left(\frac{e^{-u(t_{0})}}{t_{0}^{d/2}}\right)^{1-\epsilon } \frac{1}{t^{\epsilon d/2}}=\frac{e^{-(1-\epsilon )u(t_{0})}}{t_{0}^{d(1-\epsilon )/2}}\frac{1}{t^{\epsilon d/2}}.\) 
Now, splitting the above integral into two the integrals on \([0,1/2]\)  and \([1/2,1]\) ; we have for the first integral using (REF ),
\(\int ^{1/2}_{0}\left(-log(\sqrt{1-t})\right)^{\beta /2}\frac{e^{-(1-\epsilon )u(t)}}{t^{d/2+1}}\,dt&\lesssim &\frac{e^{-(1-\epsilon )u(t_{0})}}{t_{0}^{d(1-\epsilon )/2}}\int ^{1/2}_{0}\frac{\left(-log(\sqrt{1-t})\right)^{\beta /2}}{t^{\epsilon d/2+1}}\,dt.\) 
Set \(r=\min \lbrace \frac{\beta }{4},\frac{1}{2}\rbrace ,\,\)  then
\(0<r<1\)  and by taking \(\epsilon >0\)  such that \(\frac{\epsilon d}{2}<\frac{\beta }{2}-r\,\,\)  we get
\(\lim _{t\rightarrow 0^{+}}\frac{\left(-log(\sqrt{1-t})\right)^{\beta /2}}{t^{\epsilon d/2+r}} = \lim _{t\rightarrow 0^{+}}\left[\frac{\left(-log(\sqrt{1-t})\right)}{t}\right]^{\beta /2}t^{\beta /2 -(\epsilon d/2+r)}= 0,\) 
thus, \(\displaystyle \frac{\left(-log(\sqrt{1-t})\right)^{\beta /2}}{t^{\epsilon d/2+r}}\)  is bounded on \((0,1/2]\) ,
and hence
\(\int ^{1/2}_{0}\frac{\left(-log(\sqrt{1-t})\right)^{\beta /2}}{t^{\epsilon d/2+1}}\,dt=\int ^{1/2}_{0}\frac{\left(-log(\sqrt{1-t})\right)^{\beta /2}}{t^{\epsilon d/2+r}}\frac{dt}{t^{1-r}}\le C_{\epsilon ,\beta }\int ^{1/2}_{0}\frac{dt}{t^{1-r}}=C_{\epsilon ,\beta }.\) 
Then,
\(\int ^{1/2}_{0}\left(-log(\sqrt{1-t})\right)^{\beta /2}\frac{e^{-(1-\epsilon )u(t)}}{t^{d/2+1}}\,dt\;\le \;C_{\epsilon ,\beta }\frac{e^{-(1-\epsilon )u(t_{0})}}{t_{0}^{d(1-\epsilon )/2}}.\) 
For the integral on \([1/2,1]\)  we have, using again (REF ) and Lemma  REF
\(\int ^{1}_{1/2}\left(-log(\sqrt{1-t})\right)^{\beta /2}\frac{e^{-(1-\epsilon )u(t)}}{t^{d/2+1}}\,dt&=&\int ^{1}_{1/2}\left(-log(\sqrt{1-t})\right)^{\beta /2}\left(\frac{e^{-u(t)}}{t^{d/2}}\right)^{(1-\epsilon )}\,\frac{dt}{t^{\epsilon d/2+1}}\\&\le &C_{\epsilon }\frac{e^{-(1-\epsilon )u(t_{0})}}{t_{0}^{d(1-\epsilon )/2}}\int ^{1}_{1/2}\left(-log(\sqrt{1-t})\right)^{\beta /2}\,dt\\&\le &C_{\epsilon }\frac{e^{-(1-\epsilon )u(t_{0})}}{t_{0}^{d(1-\epsilon )/2}}\int ^{1}_{0}\left(-log(\sqrt{1-t})\right)^{\beta /2}\,dt\\&=&C_{\epsilon ,\beta }\frac{e^{-(1-\epsilon )u(t_{0})}}{t_{0}^{d(1-\epsilon )/2}}.\) 
Therefore, since \(t_{0}<1\)  and
\(\frac{d}{2}(1-\epsilon )<\frac{d}{2},\)  we get
\(II\le C_{\epsilon ,\beta }\frac{e^{-(1-\epsilon )u(t_{0})}}{t_{0}^{d(1-\epsilon )/2}}\le C_{\epsilon }\frac{e^{-(1-\epsilon )u(t_{0})}}{t_{0}^{d/2}}.\) 
Now, using (REF ) and (REF ) we get
\(III=\frac{1}{4}\int ^{1}_{0}\left(-log(\sqrt{1-t})\right)^{\beta /2}\frac{e^{-u(t)}}{t^{d/2}}\frac{dt}{t}&\lesssim &\frac{e^{-u(t_{0})}}{t_{0}^{d/2}}\int ^{1}_{0}\left(-log(\sqrt{1-t})\right)^{\beta /2}\frac{dt}{t}\\&=&C_{\beta }\frac{e^{-u(t_{0})}}{t_{0}^{d/2}}\le C_{\beta }\frac{e^{-(1-\epsilon )u(t_{0})}}{t_{0}^{d/2}}.\) 
Now, to estimate \(I\) , we use again inequality (REF )
\(I&=&\int ^{1}_{0}\left(-log(\sqrt{1-t})\right)^{\beta /2}\frac{e^{-u(t)}}{t^{d/2}}\frac{\left|y-\sqrt{1-t}x\right|}{\sqrt{t}}\frac{|x|}{\sqrt{1-t}}\frac{dt}{\sqrt{t}}\\&\le &C_{\epsilon }|x|\int ^{1}_{0}\left(-log(\sqrt{1-t})\right)^{\beta /2}\frac{e^{-(1-\epsilon )u(t)}}{t^{d/2}\sqrt{t}}\frac{dt}{\sqrt{1-t}}.\) 
Again, splitting the above integral into two the integrals on \([0,1/2]\)  and \([1/2,1]\) . For the second integral, using (REF ), that \(t \ge 1/2\)  and Lemma REF , we get
\(\int _{1/2}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )u(t)}}{t^{\frac{d}{2}}\sqrt{t}}\frac{dt}{\sqrt{1-t}}&\lesssim & e^{-(1-\epsilon )u(t_{0})}\int _{1/2}^{1}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{dt}{\sqrt{1-t}}\\&=&C_{\beta }e^{-(1-\epsilon )u(t_{0})},\) 
Next, for the integral on \([0,1/2]\)  we need to consider two cases:

Case \(\beta >0\)  and \(d=1\) : by Lemma REF  and the fact that \(\frac{-log(\sqrt{1-t})}{t}\)  is bounded on \((0, 1/2],\)
\(\int _{0}^{1/2}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )u(t)}}{t^{\frac{1}{2}}\sqrt{t}}\frac{dt}{\sqrt{1-t}}&=&\int _{0}^{1/2}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )u(t)}}{t}\frac{dt}{\sqrt{1-t}}\\&=&\int _{0}^{1/2}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}-1}\frac{\left(-log(\sqrt{1-t})\right)}{t}e^{-(1-\epsilon )u(t)}\frac{dt}{\sqrt{1-t}}\\&\le &C_{\beta }\int _{0}^{1/2}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}-1}e^{-(1-\epsilon )u(t)}\frac{dt}{\sqrt{1-t}}\\&\le &C_{\beta }e^{-(1-\epsilon )u(t_{0})}\int _{0}^{1/2}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}-1}dt\\&=&C_{\beta }e^{-(1-\epsilon )u(t_{0})}.\) 

Case \(\beta \ge 1\)  and \(d\ge 2\) : by taking \(\epsilon <\frac{2}{d}\)
\(\int _{0}^{1/2}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )u(t)}}{t^{\frac{d}{2}}\sqrt{t}}\frac{dt}{\sqrt{1-t}}&\le & C\int _{0}^{1/2}\left(-log(\sqrt{1-t})\right)^{\frac{(\beta -1)}{2}}\frac{e^{-(1-\epsilon )u(t)}}{t^{\frac{d}{2}}}\frac{dt}{\sqrt{1-t}}\\&= &C\int _{0}^{1/2}\frac{\left(-log(\sqrt{1-t})\right)^{\frac{(\beta -1)}{2}}}{t^{\frac{(d-2)}{2}}}\frac{e^{-(1-\epsilon )u(t)}}{t\sqrt{1-t}}dt\\&= &C\int _{0}^{1/2}\frac{\left(-log(\sqrt{1-t})\right)^{\frac{(\beta -1)}{2}}}{t^{\frac{d}{2}\frac{(d-2)}{d}}}e^{-(\frac{d-2}{d})u(t)}\frac{e^{(\epsilon -\frac{2}{d})u(t)}}{t\sqrt{1-t}}dt\\&\le &C\frac{e^{-(\frac{d-2}{d})u(t_{0})}}{t_{0}^{\frac{d}{2}\frac{(d-2)}{d}}}\int _{0}^{1/2}\left(-log(\sqrt{1-t})\right)^{\frac{(\beta -1)}{2}}\frac{e^{(\epsilon -\frac{2}{d})u(t)}}{t\sqrt{1-t}}dt.\) 
Since
\(\left(-log(\sqrt{1-t})\right)^{\frac{(\beta -1)}{2}}\)  is continuous
on \([0,1/2]\)  for \(\beta \ge 1\)  and proceeding in analogous way as in Lemma 4.36 of [2]}, we get
\(\int _{0}^{1/2}\left(-log(\sqrt{1-t})\right)^{\frac{\beta }{2}}\frac{e^{-(1-\epsilon )u(t)}}{t^{\frac{d}{2}}\sqrt{t}}\frac{dt}{\sqrt{1-t}} &\le & C_{\beta }\frac{e^{-(\frac{d-2}{d})u(t_{0})}}{t_{0}^{\frac{d}{2}\frac{(d-2)}{d}}}\int _{0}^{1}\frac{e^{(\epsilon -\frac{2}{d})u(t)}}{t\sqrt{1-t}}dt\\&\le &C_{\beta }\frac{e^{-(\frac{d-2}{d})u(t_{0})}}{t_{0}^{\frac{d}{2}-1}}e^{(\epsilon -\frac{2}{d})u(t_{0})}\\&= &C_{\beta }\frac{e^{-(1-\epsilon )u(t_{0})}}{t_{0}^{\frac{d}{2}-1}}= C_{\beta }\frac{e^{-(1-\epsilon )u(t_{0})}}{t_{0}^{\frac{d}{2}}}t_{0}.\) 
Thus,
\(I\le C_{\epsilon ,\beta }\displaystyle |x|\left(1+\frac{1}{t_{0}^{\frac{d}{2}}}t_{0}\right)e^{-(1-\epsilon )u(t_{0})}.\) 

Finally, since \(\displaystyle \frac{1}{t_{0}^{d/2}}\lesssim |x+y|^{d}\) , \(t_{0}<1\) , and \(|x|\le |x+y|\)  as \(b>0\) , we have

For \(|x|<1\) ,
\(I&\le & C_{\epsilon ,\beta }\displaystyle |x|\left(1+\frac{1}{t_{0}^{\frac{d}{2}}}t_{0}\right)e^{-(1-\epsilon )u(t_{0})}\le C_{\epsilon ,\beta }\frac{1}{t_{0}^{\frac{d}{2}}}e^{-(1-\epsilon )u(t_{0})}\\&\le & C_{\epsilon ,\beta }|x+y|^{d}e^{-(1-\epsilon )u(t_{0})}.\) 

For \(|x|\ge 1\) , since \(b>0\) , \(|x|\le |x+y|\)
\(I&\le & C_{\epsilon ,\beta }\displaystyle |x|\left(1+\frac{1}{t_{0}^{\frac{d}{2}}}t_{0}\right)e^{-(1-\epsilon )u(t_{0})}\\&=& C_{\epsilon ,\beta }|x|e^{-(1-\epsilon )u(t_{0})}\,+\, C_{\epsilon ,\beta }|x|t_{0}\frac{e^{-(1-\epsilon )u(t_{0})}}{t_{0}^{\frac{d}{2}}}\\&\le &C_{\epsilon ,\beta }|x+y|^{d}e^{-(1-\epsilon )u(t_{0})}\,+\, C_{\epsilon ,\beta }t_{0}|x||x+y|^{d}e^{-(1-\epsilon )u(t_{0})}.\) 

Given that \(t_{0}\le C\frac{|x-y|}{|x+y|}\)  and the fact that \(|x|\le |x+y| \)  we get, for \(|x-y|<1\) ,
\(|x|t_{0}\le C\frac{|x||x-y|}{|x+y|}\le C.\) 
Thus, for \(|x-y|<1\) ,
\(I\le C_{\epsilon ,\beta }|x+y|^{d}e^{-(1-\epsilon )u(t_{0})};\) 
and for \(|x-y|\ge 1\) ,
\(I\le C_{\epsilon ,\beta }|x+y|^{d+1}e^{-(1-\epsilon )u(t_{0})}.\) 
Hence, we conclude that

\(\left|N_{\beta /2}(x,y)\right|\le C_{\epsilon ,\beta }|x+y|^{d}e^{-(1-\epsilon )u(t_{0})},\)   for either \(|x|\le 1,\)  or for \(|x|\ge 1\)  with \(|x-y|<1\) .

\(\left|N_{\beta /2}(x,y)\right|\le C_{\epsilon ,\beta }|x+y|^{d+1}e^{-(1-\epsilon )u(t_{0})},\)   for  \(|x|\ge 1\)  with \(|x-y|\ge 1\) .

Now, as \(b>0\) , for \(f\in L^{p(\cdot )}(\mathbb {R}^{d},\gamma _{d})\)  with \(\Vert f\Vert _{p(\cdot ),\gamma _{d}}=1\) , we have that
\(&& \int _{\mathbb {R}^{d}}\left(\int \limits _{B_{h}^{c}(x)\cap E_{x}}\left|N_{\beta /2}(x,y)\right||f(y)|dy\right)^{p(x)}\gamma _{d}(dx)= \int _{|x|<1}\left(\int \limits _{B_{h}^{c}(x)\cap E_{x}}\left|N_{\beta /2}(x,y)\right||f(y)|dy\right)^{p(x)}\gamma _{d}(dx)\\&& \hspace{14.22636pt}+ \int _{|x|\ge 1}\left(\int \limits _{B_{h}^{c}(x)\cap E_{x},|x-y|<1}\left|N_{\beta /2}(x,y)\right||f(y)|dy+\int \limits _{B_{h}^{c}(x)\cap E_{x},|x-y|\ge 1}\left|N_{\beta /2}(x,y)\right||f(y)|dy\right)^{p(x)}\gamma _{d}(dx)\) 
\(&\le & \int _{|x|<1}\left(\int \limits _{B_{h}^{c}(x)\cap E_{x}}\left|N_{\beta /2}(x,y)\right||f(y)|dy\right)^{p(x)}\gamma _{d}(dx)\\&+& C\int _{|x|\ge 1}\left(\left(\int \limits _{B_{h}^{c}(x)\cap E_{x},|x-y|<1}\left|N_{\beta /2}(x,y)\right||f(y)|dy\right)^{p(x)}+\left(\int \limits _{B_{h}^{c}(x)\cap E_{x},|x-y|\ge 1}\left|N_{\beta /2}(x,y)\right||f(y)|dy\right)^{p(x)}\right)\gamma _{d}(dx)\\&\le &C_{\epsilon ,\beta }\int _{R^{d}}\left(\int \limits _{B_{h}^{c}(x)\cap E_{x}}|x+y|^{d}e^{-(1-\epsilon )u(t_{0})}|f(y)|dy\right)^{p(x)}\gamma _{d}(dx)\\&& \hspace{21.33955pt}+ C_{\epsilon ,\beta }\int _{R^{d}}\left(\int \limits _{B_{h}^{c}(x)\cap E_{x},|x-y|\ge 1}|x+y|^{d+1}e^{-(1-\epsilon )u(t_{0})}|f(y)|dy\right)^{p(x)}\gamma _{d}(dx)\\&=&C_{\epsilon ,\beta }\int _{R^{d}}\left(\int \limits _{B_{h}^{c}(x)\cap E_{x}}|x+y|^{d}e^{-(1-\epsilon )u(t_{0})}e^{|y|^{2}/p(y)-|x|^{2}/p(x)}|f(y)|e^{-|y|^{2}/p(y)}dy\right)^{p(x)}dx\\&& \hspace{21.33955pt}+ C_{\epsilon ,\beta }\int _{R^{d}}\left(\int \limits _{B_{h}^{c}(x)\cap E_{x},|x-y|\ge 1}|x+y|^{d+1}e^{-(1-\epsilon )u(t_{0})}e^{|y|^{2}/p(y)-|x|^{2}/p(x)}|f(y)|e^{-|y|^{2}/p(y)}dy\right)^{p(x)}dx.\) 
Since \(p(\cdot )\in \mathcal {P}_{\gamma _{d}}^{\infty }(\mathbb {R}^{d})\) , we obtain that \(e^{|y|^{2}/p(y)-|x|^{2}/p(x)}\approx e^{(|y|^{2}-|x|^{2})/p_{\infty }},\)  and by the Cauchy-Schwartz inequality we have,
\(\left| |y|^{2}-|x|^{2}\right|\le |x+y| |x-y|,\) 
for all \(x,y\in \mathbb {R}^{d}\) . Therefore,
\(&&\int _{B^{c}_h(x)\cap E_{x}}|x+y|^{d}e^{-(1-\epsilon )u(t_{0})}e^{|y|^{2}/p(y)-|x|^{2}/p(x)}|f(y)|e^{-|y|^{2}/p(y)}dy \hspace{99.58464pt}\\&& \hspace{184.9429pt} \lesssim \int _{B^{c}_h(x)\cap E_{x}}P(x,y)|f(y)|e^{-|y|^{2}/p(y)}dy,\) 
and
\(&&\int _{B^{c}_h(x)\cap E_{x},|x-y|\ge 1}|x+y|^{d+1}e^{-(1-\epsilon )u(t_{0})}e^{|y|^{2}/p(y)-|x|^{2}/p(x)}|f(y)|e^{-|y|^{2}/p(y)}dy \hspace{99.58464pt}\\&& \hspace{156.49014pt} \lesssim \int _{B^{c}_h(x)\cap E_{x}}Q(x,y)|f(y)|e^{-|y|^{2}/p(y)}dy,\) 
where
\(P(x,y)=|x+y|^{d}e^{-\alpha _{\infty }|x+y||x-y|}, \; Q(x,y)=|x+y|^{d+1}e^{-\alpha _{\infty }|x+y|}\) 
and
\( \alpha _{\infty }=\left(\frac{1-\epsilon }{2}-\left|\frac{1}{p_{\infty }}-\frac{1-\epsilon }{2}\right|\right).\) 
It is easy to see that \(\alpha _{\infty }>0\)  if \(\epsilon <1/p^{\prime }_{\infty }\) .
Therefore, in order to make sense of all the estimates above we need to take
\(0<\epsilon <\displaystyle \min \left\lbrace \frac{1}{d},\frac{1}{p^{\prime }_{\infty }},\frac{2}{d}\left(\frac{\beta }{2}-r\right)\right\rbrace .\) 
Observe that \(P(x,y)\)  is the same kernel considered in the proof of Theorem 3.5, page 416 of [12]}, so we can conclude that
\(\displaystyle \int _{R^{d}}\left(\int _{B^{c}_h(x)\cap E_{x}}P(x,y)|f(y)|e^{-|y|^{2}/p(y)}dy\right)^{p(x)}dx\le C.\) 
On the other hand, it can be proved that \(Q(x,y)\)  is integrable on each variable and the value of each integral is independent of \(x\)  and \(y\) .
Now, we use an analogous argument as in [12]} for \(Q(x,y)\) . Taking,
\(J=\displaystyle \int _{R^{d}}Q(x,y)|f(y)|e^{-|y|^{2}/p(y)}dy,\) 
and using Hölder's inequality, we obtain
\( J\lesssim \Vert Q(x,\cdot )\Vert _{p^{\prime }(\cdot )}\Vert fe^{-|\cdot |^{2}/p(\cdot )}\Vert _{p(\cdot )}\le \Vert Q(x,\cdot )\Vert _{p^{\prime }(\cdot )}.\) 
and,
\(\int _{R^{d}}Q(x,y)^{p^{\prime }(y)}dy&=&\int _{R^{d}}|x+y|^{(d+1)p^{\prime }(y)}e^{-\alpha _{\infty }|x+y|p^{\prime }(y)}dy\\&\le &\int _{|x+y|<1}|x+y|^{d+1}e^{-\alpha _{\infty }|x+y|}dy +\int _{|x+y|\ge 1}|x+y|^{(d+1)p^{\prime }_{+}}e^{-\alpha _{\infty }|x+y|}dy\\&\le &\int _{R^{d}}\left(|z|^{d+1}+|z|^{(d+1)p^{\prime }_{+}}\right)e^{-\alpha _{\infty }|z|}dz\\&=&C_{p,d}.\) 
Thus, \(J\lesssim \Vert Q(x,\cdot )\Vert _{p^{^{\prime }}(\cdot )}\le C_{p,d}\) , and therefore
\(\frac{1}{C_{p,d}}\int _{R^{d}}Q(x,y)|f(y)|e^{-|y|^{2}/p(y)}dy\le 1.\) 
We set \(g(y)=|f(y)|e^{-|y|^{2}/p(y)}=g_{1}(y)+g_{2}(y)\) , where \(g_{1}=g\chi _{\lbrace g\ge 1\rbrace }\)  and \(g_{2}=g\chi _{\lbrace g<1\rbrace }\) , then
\(\int _{\mathbb {R}^{d}}J^{p(x)}dx&=& \int _{\mathbb {R}^{d}}\left(\int _{\mathbb {R}^{d}}Q(x,y)|f(y)|e^{-|y|^{2}/p(y)}dy\right)^{p(x)}dx \\&\lesssim & \int _{\mathbb {R}^{d}}\left(\frac{1}{C_{p,d}}\int _{\mathbb {R}^{d}}Q(x,y)g(y)dy\right)^{p(x)}dx\\&\lesssim & \int _{\mathbb {R}^{d}}\left(\frac{1}{C_{p,d}}\int _{\mathbb {R}^{d}}Q(x,y)g_{1}(y)dy\right)^{p(x)}dx+\int _{\mathbb {R}^{d}}\left(\frac{1}{C_{p,d}}\int _{\mathbb {R}^{d}}Q(x,y)g_{2}(y)dy\right)^{p(x)}dx.\) 
By Hölder's inequality and Fubini's theorem
\(\int _{\mathbb {R}^{d}}\left(\frac{1}{C_{p,d}}\int _{\mathbb {R}^{d}}Q(x,y)g_{1}(y)dy\right)^{p(x)}dx&\lesssim &\int _{\mathbb {R}^{d}}\left(\int _{\mathbb {R}^{d}}Q(x,y)g_{1}(y)dy\right)^{p_{-}}dx\\&=&\int _{\mathbb {R}^{d}}\left(\int _{\mathbb {R}^{d}}Q^{\frac{1}{p^{\prime }_{-}}}(x,y)Q^{\frac{1}{p_{-}}}(x,y)g_{1}(y)dy\right)^{p_{-}}dx\\&\le &\int _{\mathbb {R}^{d}}\left(\int _{\mathbb {R}^{d}}Q(x,y)dy\right)^{p_{-}/p^{\prime }_{-}}\left(\int _{\mathbb {R}^{d}}Q(x,y)g^{p_{-}}_{1}(y)dy\right)dx\\&=&C_{p}\int _{\mathbb {R}^{d}}\left(\int _{\mathbb {R}^{d}}Q(x,y)g^{p_{-}}_{1}(y)dy\right)dx\\&=&C_{p}\int _{\mathbb {R}^{d}}g^{p_{-}}_{1}(y)\left(\int _{\mathbb {R}^{d}}Q(x,y)dx\right)dy=C_{p}\int _{\mathbb {R}^{d}}g^{p_{-}}_{1}(y)dy.\) 
Therefore,
\(\int _{\mathbb {R}^{d}}\left(\frac{1}{C_{p,d}}\int _{\mathbb {R}^{d}}Q(x,y)g_{1}(y)dy\right)^{p(x)}dx&\lesssim &\int _{\mathbb {R}^{d}}g^{p(y)}_{1}(y)dy\le \int _{\mathbb {R}^{d}}|f(y)|^{p(y)}e^{-|y|^{2}}dy\\&\lesssim & \rho _{p(\cdot ),\gamma _{d}}(f)\le 1.\) 
On the other hand, applying the inequality (REF ) in Lemma REF , since
\(G(x):=\displaystyle \frac{1}{C_{p,d}}\int _{\mathbb {R}^{d}}Q(x,y)g_{2}(y)dy\le 1\) , we obtain that
\(\int _{\mathbb {R}^{d}}\left(\int _{\mathbb {R}^{d}}\frac{1}{C_{p,d}}Q(x,y)g_{2}(y)dy\right)^{p(x)}dx&=&\int _{\mathbb {R}^{d}}(G(x))^{p(x)}dx \\&\le & \int _{\mathbb {R}^{d}}(G(x))^{p_{\infty }}dx + \int _{\mathbb {R}^{d}}\frac{dx}{(e+|x|)^{dp_{-}}}\\&\lesssim &\int _{\mathbb {R}^{d}}\left(\int _{\mathbb {R}^{d}}Q(x,y)g_{2}(y)dy\right)^{p_{\infty }}+C_{d,p}.\) 
Finally, to estimate the integral
\(\int _{\mathbb {R}^{d}}\left(\int _{\mathbb {R}^{d}}Q(x,y)g_{2}(y)dy\right)^{p_{\infty }}dx,\) 
we proceed in an analogous way, applying Hölder's inequality to the exponent \(p_{\infty }\) , Fubbini's theorem and inequality () in Lemma REF , we get
\(\int _{\mathbb {R}^{d}}\left(\int _{\mathbb {R}^{d}}Q(x,y)g_{2}(y)dy\right)^{p_{\infty }}dx &\lesssim &\int _{\mathbb {R}^{d}}g_{2}^{p_{\infty }}(y)dy\\&\le &\int _{\mathbb {R}^{d}}g_{2}^{p(y)}(y)dy+\int _{\mathbb {R}^{d}}\frac{dy}{(e+|y|)^{dp_{-}}}\\&\le &\int _{\mathbb {R}^{d}}|f(y)|^{p(y)}e^{-|y|^{2}}dy+C_{d,p}\lesssim 1+C_{d,p}.\) 
Thus,
\(\int _{\mathbb {R}^{d}}\left(\int \limits _{B_{h}^{c}(x)\cap E_{x}}\left|N_{\beta /2}(x,y)\right||f(y)|dy\right)^{p(x)}\gamma _{d}(dx)&\le & C_{d,p}.\) 
With this, we obtain that \(\Vert I_{\beta }(f\chi _{B_{h}^{c}(\cdot )\cap E_{(\cdot )}})\Vert _{p(\cdot ),\gamma _{d}}\le C_{d,p}\) .
We conclude that
\(\left\Vert I_{\beta ,G}(f)\right\Vert _{p(\cdot ), \gamma _{d}}=\left\Vert I_{\beta }\left(f \chi _{B^{c}_h(\cdot )}\right)\right\Vert _{p(\cdot ), \gamma _{d}} \le C,\) 
and by homogeneity of the norm,
\(\left\Vert I_{\beta ,G}(f)\right\Vert _{p(\cdot ), \gamma _{d}} \le C\Vert f\Vert _{p(\cdot ),\gamma _{d}}\) , for all function \(f\in L^{p(\cdot )}(\mathbb {R}^{d},\gamma _{d})\) .
Therefore, the global part \(I_{\beta ,G}\)  is bounded in \(L^{p(\cdot )}(\gamma _d)\)  and the proof is complete.


